3.145 \(\int e^{\frac{3}{2} i \tan ^{-1}(a x)} x^m \, dx\)

Optimal. Leaf size=36 \[ \frac{x^{m+1} F_1\left (m+1;\frac{3}{4},-\frac{3}{4};m+2;i a x,-i a x\right )}{m+1} \]

[Out]

(x^(1 + m)*AppellF1[1 + m, 3/4, -3/4, 2 + m, I*a*x, (-I)*a*x])/(1 + m)

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Rubi [A]  time = 0.0260395, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {5062, 133} \[ \frac{x^{m+1} F_1\left (m+1;\frac{3}{4},-\frac{3}{4};m+2;i a x,-i a x\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[E^(((3*I)/2)*ArcTan[a*x])*x^m,x]

[Out]

(x^(1 + m)*AppellF1[1 + m, 3/4, -3/4, 2 + m, I*a*x, (-I)*a*x])/(1 + m)

Rule 5062

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[(x^m*(1 - I*a*x)^((I*n)/2))/(1 + I*a*x)^((I*n)/2
), x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[(I*n - 1)/2]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int e^{\frac{3}{2} i \tan ^{-1}(a x)} x^m \, dx &=\int \frac{x^m (1+i a x)^{3/4}}{(1-i a x)^{3/4}} \, dx\\ &=\frac{x^{1+m} F_1\left (1+m;\frac{3}{4},-\frac{3}{4};2+m;i a x,-i a x\right )}{1+m}\\ \end{align*}

Mathematica [F]  time = 0.182232, size = 0, normalized size = 0. \[ \int e^{\frac{3}{2} i \tan ^{-1}(a x)} x^m \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[E^(((3*I)/2)*ArcTan[a*x])*x^m,x]

[Out]

Integrate[E^(((3*I)/2)*ArcTan[a*x])*x^m, x]

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Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) ^{{\frac{3}{2}}}{x}^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)*x^m,x)

[Out]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)*x^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m} \left (\frac{i \, a x + 1}{\sqrt{a^{2} x^{2} + 1}}\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)*x^m,x, algorithm="maxima")

[Out]

integrate(x^m*((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{i \, \sqrt{a^{2} x^{2} + 1} x^{m} \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}}{a x + i}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)*x^m,x, algorithm="fricas")

[Out]

integral(I*sqrt(a^2*x^2 + 1)*x^m*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))/(a*x + I), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a**2*x**2+1)**(1/2))**(3/2)*x**m,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m} \left (\frac{i \, a x + 1}{\sqrt{a^{2} x^{2} + 1}}\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)*x^m,x, algorithm="giac")

[Out]

integrate(x^m*((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(3/2), x)